Post on 01-Apr-2015
© 2003-2008BYU
07 KMPage 1
ECEn/CS 224
Karnaugh Maps
© 2003-2008BYU
07 KMPage 2
ECEn/CS 224
What are Karnaugh Maps?
A simpler way to handle most (but not all) jobs of manipulating logic functions.
Hooray !!
© 2003-2008BYU
07 KMPage 3
ECEn/CS 224
Karnaugh Map Advantages
• Minimization can be done more systematically
• Much simpler to find minimum solutions
• Easier to see what is happening (graphical)
Almost always used instead
of boolean minimization.
© 2003-2008BYU
07 KMPage 4
ECEn/CS 224
Gray Codes
• Gray code is a binary value encoding in which adjacent values only differ by one bit
2-bit Gray Code
00
01
11
10
© 2003-2008BYU
07 KMPage 5
ECEn/CS 224
Truth Table Adjacencies
A B F
0 0 10 1 11 0 01 1 0
F = A’
These are adjacent in a gray code sense - they differ by 1 bit
We can apply XY + XY’ = X
A’B’ + A’B = A’(B’+B) = A’(1) = A’A B F
0 0 00 1 11 0 01 1 1
F = B Same idea:
A’B + AB = B
Key idea:
Gray code adjacency allows use of simplification theorems
Problem:
Physical adjacency in truth table does not indicate gray code adjacency
© 2003-2008BYU
07 KMPage 6
ECEn/CS 224
2-Variable Karnaugh Map
A=0, B=0
A=0, B=1
A=1, B=0
A=1, B=1
A different way to draw a truth table: by folding it
A
B 0 1
0
1
A B F
0 0
0 1
1 0
1 1
© 2003-2008BYU
07 KMPage 7
ECEn/CS 224
Karnaugh Map
• In a K-map, physical adjacency does imply gray code adjacency
A
B 0 1
0 0 0
1 1 1
F = A’B + AB = B
A
B 0 1
0 1 0
1 1 0
F =A’B’ + A’B = A’
© 2003-2008BYU
07 KMPage 8
ECEn/CS 224
2-Variable Karnaugh Map
A B F
0 0 10 1 11 0 01 1 0
© 2003-2008BYU
07 KMPage 9
ECEn/CS 224
2-Variable Karnaugh Map
A B F
0 0 10 1 11 0 01 1 0
A
B 0 1
0
1
© 2003-2008BYU
07 KMPage 10
ECEn/CS 224
2-Variable Karnaugh Map
A B F
0 0 10 1 11 0 01 1 0
A
B 0 1
0 1
1 1
© 2003-2008BYU
07 KMPage 11
ECEn/CS 224
2-Variable Karnaugh Map
A
B 0 1
0 1 0
1 1 0
A B F
0 0 10 1 11 0 01 1 0
© 2003-2008BYU
07 KMPage 12
ECEn/CS 224
2-Variable Karnaugh Map
F = A’B’ + A’B = A’
A B F
0 0 10 1 11 0 01 1 0
A
B 0 1
0 1 0
1 1 0
© 2003-2008BYU
07 KMPage 13
ECEn/CS 224
2-Variable Karnaugh Map
A B F
0 0 10 1 11 0 01 1 0
A
B 0 1
0 1 0
1 1 0
A = 0
F = A’
© 2003-2008BYU
07 KMPage 14
ECEn/CS 224
A
B 0 1
0 0 1
1 1 1
Another Example
F = A’B + AB’ + AB = (A’B + AB) + (AB’ + AB) = A + B
A B F
0 0 00 1 11 0 11 1 1
© 2003-2008BYU
07 KMPage 15
ECEn/CS 224
A
B 0 1
0 0 1
1 1 1
F = A + B
Another Example
A B F
0 0 00 1 11 0 11 1 1 A = 1
B = 1
© 2003-2008BYU
07 KMPage 16
ECEn/CS 224
A
B 0 1
0 1 1
1 1 1
Yet Another Example
F = 1
A B F
0 0 10 1 11 0 11 1 1
Groups of more than two 1’s can be combined
© 2003-2008BYU
07 KMPage 17
ECEn/CS 224
3-Variable Karnaugh Map Showing Minterm Locations
Note the order ofthe B C variables:
0 0 0 1 1 1 1 0
A
BC 0 1
00
01
11
10
ABC = 010
ABC = 101
© 2003-2008BYU
07 KMPage 18
ECEn/CS 224
3-Variable Karnaugh Map Showing Minterm Locations
ABC = 010
ABC = 101
Note the order ofthe B C variables:
0 0 0 1 1 1 1 0
A
BC 0 1
00 m0 m4
01 m1 m5
11 m3 m7
10 m2 m6
© 2003-2008BYU
07 KMPage 19
ECEn/CS 224
Adjacencies
• Adjacent squares differ by exactly one variable
A
BC 0 1
00 AB'C'
01 A'B'C AB'C
11 ABC
10 ABC'
There is wrap-around: top and bottom rows are adjacent
© 2003-2008BYU
07 KMPage 20
ECEn/CS 224
Truth Table to Karnaugh Map
A
BC 0 1
00
01
11
10
A B C F
0 0 0 00 0 1 00 1 0 10 1 1 11 0 0 01 0 1 11 1 0 01 1 1 1
0
0
1
1
0
1
1
0
© 2003-2008BYU
07 KMPage 21
ECEn/CS 224
A
BC 0 1
00 0 0
01 0 1
11 1 1
10 1 0
Minimization Example
A’BC+A’BC’ = A’B
AB’C+ABC = AC
F = A’B + AC
© 2003-2008BYU
07 KMPage 22
ECEn/CS 224
Another Example
A’B’C+A’BC = A’C AB’C’+ABC’ = AC’
F = A’C + AC’ = A C
A
BC 0 1
00 0 1
01 1 0
11 1 0
10 0 1
© 2003-2008BYU
07 KMPage 23
ECEn/CS 224
Minterm Expansion to K-Map
F = m( 1, 3, 4, 6 )
A
BC 0 1
00
01
11
10
A
BC 0 1
00 m0 m4
01 m1 m5
11 m3 m7
10 m2 m6
Minterms are the 1’s, everything else is 0
0
1
1
0
1
0
0
1
© 2003-2008BYU
07 KMPage 24
ECEn/CS 224
Maxterm Expansion to KMap
A
BC 0 1
00
01
11
10
A
BC 0 1
00 M0 M4
01 M1 M5
11 M3 M7
10 M2 M6
F = ПM( 0, 2, 5, 7 )
Maxterms are the 0’s, everything else is 1
0
1
1
0
1
0
0
1
© 2003-2008BYU
07 KMPage 25
ECEn/CS 224
Yet Another Example
A’B’C’+AB’C’+A’B’C+AB’C = B’
AB’C+ABC = AC
F = B’ + AC
2n 1’s can be circled at a time1, 2, 4, 8, … OK
3 not OK
The larger the group of 1’s the simpler the resulting product term
A
BC 0 1
00 1 1
01 1 1
11 0 1
10 0 0
© 2003-2008BYU
07 KMPage 26
ECEn/CS 224
A
BC 0 1
00
01
11
10
Boolean Algebra to Karnaugh Map
Plot: ab’c’ + bc + a’
© 2003-2008BYU
07 KMPage 27
ECEn/CS 224
Boolean Algebra to Karnaugh Map
A
BC 0 1
00 1
01
11
10
Plot: ab’c’ + bc + a’
© 2003-2008BYU
07 KMPage 28
ECEn/CS 224
A
BC 0 1
00 1
01
11 1 1
10
Boolean Algebra to Karnaugh Map
Plot: ab’c’ + bc + a’
© 2003-2008BYU
07 KMPage 29
ECEn/CS 224
Boolean Algebra to Karnaugh Map
Plot: ab’c’ + bc + a’
A
BC 0 1
00 1 1
01 1
11 1 1
10 1
© 2003-2008BYU
07 KMPage 30
ECEn/CS 224
A
BC 0 1
00 1 1
01 1 0
11 1 1
10 1 0
Boolean Algebra to Karnaugh Map
Remainingspaces are 0
Plot: ab’c’ + bc + a’
© 2003-2008BYU
07 KMPage 31
ECEn/CS 224
A
BC 0 1
00 1 1
01 1 0
11 1 1
10 1 0
Boolean Algebra to Karnaugh Map
F = B’C’ + BC + A’
This is a simpler equation thanwe started with.
Do you see how we obtained it?
Now minimize . . .
© 2003-2008BYU
07 KMPage 32
ECEn/CS 224
Mapping Sum of Product Terms
The 3-variable map has 12 possible groups of 2 spaces
These become terms with 2 literals
A
BC 0 1
00
01
11
10
A
BC 0 1
00
01
11
10
A
BC 0 1
00
01
11
10
© 2003-2008BYU
07 KMPage 33
ECEn/CS 224
A
BC 0 1
00
01
11
10
A
BC 0 1
00
01
11
10
Mapping Sum of Product Terms
The 3-variable map has 6 possible groups of 4 spaces
These become terms with 1 literal
A
BC 0 1
00
01
11
10
© 2003-2008BYU
07 KMPage 34
ECEn/CS 224
AB
CD 00 01 11 10
00 0 0 0 1
01 1 1 1 1
11 1 1 1 1
10 0 1 0 0
4-Variable Karnaugh Map
Note the rowand column orderings.
Required for adjacency
AB
CD 00 01 11 10
00 m0 m4 m12 m8
01 m1 m5 m13 m9
11 m3 m7 m15 m11
10 m2 m6 m14 m10
DA’BC
AB’C’
F = A’BC + AB’C’ + D
© 2003-2008BYU
07 KMPage 35
ECEn/CS 224
AB
CD 00 01 11 10
00 1 0 0 1
01 0 0 0 0
11 1 1 1 1
10 1 1 1 0
Find a POS Solution
Find solutions to groups of 0’s to find F’Invert to get F then use DeMorgan’s
C’D
F’ = C’D + BC’ + AB’CD’F = (C+D’)(B’+C)(A’+B+C’+D)
AB’CD’
BC’
© 2003-2008BYU
07 KMPage 36
ECEn/CS 224
A
BC 0 1
00
01
11
10
Dealing With Don’t Cares
F = m(1, 3, 7) + d(0, 5)
© 2003-2008BYU
07 KMPage 37
ECEn/CS 224
A
BC 0 1
00 x 0
01 1 x
11 1 1
10 0 0
Dealing With Don’t Cares
A’B’C+AB’C+A’BC+ABC = C
F = C
Circle the x’s that help get bigger groups of 1’s (or 0’s if POS)Don’t circle the x’s that don’t
F = m(1, 3, 7) + d(0, 5)
© 2003-2008BYU
07 KMPage 38
ECEn/CS 224
Minimal K-Map Solutions
Some Terminologyand
An Algorithm to Find Them
© 2003-2008BYU
07 KMPage 39
ECEn/CS 224
Prime Implicants
• A group of one or more 1’s which are adjacent and can be combined on a Karnaugh Map is called an implicant.
• The biggest group of 1’s which can be circled to cover a given 1 is called a prime implicant.
– They are the only implicants we care about.
© 2003-2008BYU
07 KMPage 40
ECEn/CS 224
Prime Implicants
Prime Implicants Non-prime Implicants
Are there any additional prime implicants in the map that are not shown above?
AB
CD 00 01 11 10
00 0 0 0 1
01 0 0 1 1
11 0 1 1 1
10 0 1 1 1
© 2003-2008BYU
07 KMPage 41
ECEn/CS 224
All The Prime Implicants
Prime Implicants
When looking for a minimal solution – only circle prime implicants…
A minimal solution will never contain non-prime implicants
AB
CD 00 01 11 10
00 0 0 0 1
01 0 0 1 1
11 0 1 1 1
10 0 1 1 1
© 2003-2008BYU
07 KMPage 42
ECEn/CS 224
Essential Prime Implicants
Not all prime implicants are required…
A prime implicant which is the only cover of some 1 is essential – a minimal solution requires it.
Essential Prime Implicants Non-essential Prime Implicants
AB
CD 00 01 11 10
00 0 0 0 1
01 0 0 1 1
11 0 1 1 1
10 0 1 1 1
© 2003-2008BYU
07 KMPage 43
ECEn/CS 224
A Minimal Solution Example
Minimum
Not required…
AB
CD 00 01 11 10
00 0 0 0 1
01 0 0 1 1
11 0 1 1 1
10 0 1 1 1
F = AB’ + BC + AD
© 2003-2008BYU
07 KMPage 44
ECEn/CS 224
AB
CD 00 01 11 10
00 1 0 0 1
01 1 1 0 0
11 1 1 1 0
10 1 0 0 1
Another Example
© 2003-2008BYU
07 KMPage 45
ECEn/CS 224
Another Example
Minimum
A’B’ is not required…
Every one one of itslocations is covered by multiple implicants
After choosing essentials, everything is covered…
AB
CD 00 01 11 10
00 1 0 0 1
01 1 1 0 0
11 1 1 1 0
10 1 0 0 1
F = A’D + BCD + B’D’
© 2003-2008BYU
07 KMPage 46
ECEn/CS 224
Finding the Minimum Sum of Products
1. Find each essential prime implicant and include it in the solution.
2. Determine if any minterms are not yet covered.
3. Find the minimal # of remaining prime implicants which finish the cover.
© 2003-2008BYU
07 KMPage 47
ECEn/CS 224
AB
CD 00 01 11 10
00 1 1 0 0
01 0 0 1 1
11 1 1 1 1
10 1 1 0 0
Yet Another Example(Use of non-essential primes)
© 2003-2008BYU
07 KMPage 48
ECEn/CS 224
Yet Another Example(Use of non-essential primes)
AD
CDA’C
A’D’
AB
CD 00 01 11 10
00 1 1 0 0
01 0 0 1 1
11 1 1 1 1
10 1 1 0 0
Essentials: A’D’ and ADNon-essentials: A’C and CDSolution: A’D’ + AD + A’C or A’D’ + AD + CD
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07 KMPage 49
ECEn/CS 224
K-Map Solution Summary
• Identify prime implicants
• Add essentials to solution
• Find a minimum # non-essentials required to cover rest of map
© 2003-2008BYU
07 KMPage 50
ECEn/CS 224
5- and 6-Variable K-Maps
© 2003-2008BYU
07 KMPage 51
ECEn/CS 224
5-Variable Karnaugh Map
The planes are adjacent to one another (one is above the other in 3D)
BC
DE 00 01 11 10
00 m0 m4 m12 m8
01 m1 m5 m13 m9
11 m3 m7 m15 m11
10 m2 m6 m14 m10
BC
DE 00 01 11 10
00 m16 m20 m28 m24
01 m17 m21 m29 m25
11 m19 m23 m31 m27
10 m18 m22 m30 m26
This is the A=0 plane This is the A=1 plane
© 2003-2008BYU
07 KMPage 52
ECEn/CS 224
D’E’
Some Implicants in a 5-Variable KMap
A=0 A=1
AB’C’DA’BCD
B’C’DE’
ABC’DE
Some of these are not prime…
BC
DE 00 01 11 10
00 1 1 1 1
01 0 0 0 0
11 1 0 0 1
10 1 0 0 0
BC
DE 00 01 11 10
00 1 1 1 1
01 0 0 0 0
11 0 0 1 0
10 1 0 1 0
© 2003-2008BYU
07 KMPage 53
ECEn/CS 224
BC
DE 00 01 11 10
00
01
11
10
BC
DE 00 01 11 10
00
01
11
10
5-Variable KMap Example
A=0 A=1
Find the minimum sum-of-products for:F = m (0,1,4,5,11,14,15,16,17,20,21,30,31)
© 2003-2008BYU
07 KMPage 54
ECEn/CS 224
5-Variable KMap Example
A=0 A=1
Find the minimum sum-of-products for:F = m (0,1,4,5,11,14,15,16,17,20,21,30,31)
BC
DE 00 01 11 10
00 1 1 0 0
01 1 1 0 0
11 0 0 1 0
10 0 0 1 0
BC
DE 00 01 11 10
00 1 1 0 0
01 1 1 0 0
11 0 0 1 1
10 0 0 1 0
F = B’D’ + BCD + A’BDE
© 2003-2008BYU
07 KMPage 55
ECEn/CS 224
6-Variable Karnaugh MapCD
EF 00 01 11 10
00 m0 m4 m12 m8
01 m1 m5 m13 m9
11 m3 m7 m15 m11
10 m2 m6 m14 m10
AB=00
CD
EF 00 01 11 10
00 m32 m36 m44 m40
01 m33 m37 m45 m41
11 m35 m39 m47 m43
10 m34 m38 m46 m42
CD
EF 00 01 11 10
00 m16 m20 m28 m24
01 m17 m21 m29 m25
11 m19 m23 m31 m27
10 m18 m22 m30 m26
CD
EF 00 01 11 10
00 m48 m52 m60 m56
01 m49 m53 m61 m57
11 m51 m55 m63 m59
10 m50 m54 m62 m58
AB=01
AB=10
AB=11
© 2003-2008BYU
07 KMPage 56
ECEn/CS 224
CD
EF 00 01 11 10
00 0 0 0 0
01 0 0 0 0
11 0 0 1 0
10 0 0 0 0
CD
EF 00 01 11 10
00 1 0 0 0
01 1 0 0 0
11 1 0 1 0
10 1 0 0 0
CD
EF 00 01 11 10
00 1 0 0 0
01 1 0 0 0
11 1 0 1 0
10 1 0 0 0
CD
EF 00 01 11 10
00 0 0 0 0
01 0 0 0 0
11 0 0 1 0
10 0 0 0 0
AB=00
AB=01
AB=10
AB=11
= CDEF= AC’D’Solution = AC’D’ + CDEF
© 2003-2008BYU
07 KMPage 57
ECEn/CS 224
KMap Summary
• A Kmap is simply a folded truth table– where physical adjacency implies logical
adjacency
• KMaps are most commonly used hand method for logic minimization
• KMaps have other uses for visualizing Boolean equations– you may see some later.